Blog: `int root(3)(x)/(root(3)(x) - 1) dx` Find the indefinite integral by u substitution. (let u be the denominator of the integral)

Solving indefinite integral by u-substitution, we follow:

$\displaystyle\int{f{{\left({g{{\left({x}\right)}}}\right)}}}\cdot{g{'}}{\left({x}\right)}=\int{f{{\left({u}\right)}}}\cdot{d}{u}$ where we let $\displaystyle{u}={g{{\left({x}\right)}}}$ .

By following the instruction to let "u" be the denominator of the integral,

it means we let: u = $\displaystyle{\sqrt[{{3}}]{{{x}}}}-{1}$

Find the derivative of "u" which is $\displaystyle{d}{u}=\frac{{1}}{{{3}{{x}}^{{\frac{{2}}{{3}}}}}}{\left.{d}{x}\right.}$

Then $\displaystyle{d}{u}=\frac{{1}}{{{3}{{x}}^{{\frac{{2}}{{3}}}}}}{\left.{d}{x}\right.}$ can be rearrange into $\displaystyle{3}{{x}}^{{\frac{{2}}{{3}}}}{d}{u}={\left.{d}{x}\right.}$ .

Applying u-substitution using $\displaystyle{u}={\sqrt[{{3}}]{{{x}}}}-{1}$ and $\displaystyle{3}{{x}}^{{\frac{{2}}{{3}}}}{d}{u}={\left.{d}{x}\right.}$ .

$\displaystyle\int\frac{{\sqrt[{{3}}]{{{x}}}}}{{{\sqrt[{{3}}]{{{x}}}}-{1}}}{\left.{d}{x}\right.}=\int\frac{{\sqrt[{{3}}]{{{x}}}}}{{u}}\cdot{3}{{x}}^{{\frac{{2}}{{3}}}}{d}{u}$

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Solving indefinite integral by u-substitution, we follow:

$\displaystyle\int{f{{\left({g{{\left({x}\right)}}}\right)}}}\cdot{g{'}}{\left({x}\right)}=\int{f{{\left({u}\right)}}}\cdot{d}{u}$ where we let $\displaystyle{u}={g{{\left({x}\right)}}}$ .

By following the instruction to let "u" be the denominator of the integral,

it means we let: u = $\displaystyle{\sqrt[{{3}}]{{{x}}}}-{1}$

Find the derivative of "u" which is $\displaystyle{d}{u}=\frac{{1}}{{{3}{{x}}^{{\frac{{2}}{{3}}}}}}{\left.{d}{x}\right.}$

Then $\displaystyle{d}{u}=\frac{{1}}{{{3}{{x}}^{{\frac{{2}}{{3}}}}}}{\left.{d}{x}\right.}$ can be rearrange into $\displaystyle{3}{{x}}^{{\frac{{2}}{{3}}}}{d}{u}={\left.{d}{x}\right.}$ .

Applying u-substitution using $\displaystyle{u}={\sqrt[{{3}}]{{{x}}}}-{1}$ and $\displaystyle{3}{{x}}^{{\frac{{2}}{{3}}}}{d}{u}={\left.{d}{x}\right.}$ .

$\displaystyle\int\frac{{\sqrt[{{3}}]{{{x}}}}}{{{\sqrt[{{3}}]{{{x}}}}-{1}}}{\left.{d}{x}\right.}=\int\frac{{\sqrt[{{3}}]{{{x}}}}}{{u}}\cdot{3}{{x}}^{{\frac{{2}}{{3}}}}{d}{u}$

$\displaystyle=\int\frac{{{{x}}^{{\frac{{1}}{{3}}}}\cdot{3}{{x}}^{{\frac{{2}}{{3}}}}}}{{u}}{d}{u}$

$\displaystyle=\int\frac{{{3}{{x}}^{{\frac{{1}}{{3}}+\frac{{2}}{{3}}}}}}{{u}}{d}{u}$

$\displaystyle={3}\int\frac{{x}}{{u}}{d}{u}$

Note: $\displaystyle{{x}}^{{\frac{{1}}{{3}}+\frac{{2}}{{3}}}}={{x}}^{{\frac{{3}}{{3}}}}$

$\displaystyle={{x}}^{{1}}{\quad\text{or}\quad}{x}$

Algebraic techniques:

From $\displaystyle{u}={\sqrt[{{3}}]{{{x}}}}-{1}$ , we can rearrange it into $\displaystyle{\sqrt[{{3}}]{{{x}}}}={u}+{1}$ .

Raising both sides by a power 3:

$\displaystyle{{\left({\sqrt[{{3}}]{{{x}}}}\right)}}^{{3}}={{\left({u}+{1}\right)}}^{{3}}$

$\displaystyle{x}={\left({u}+{1}\right)}\cdot{\left({u}+{1}\right)}\cdot{\left({u}+{1}\right)}$

By FOIL: $\displaystyle{\left({u}+{1}\right)}\cdot{\left({u}+{1}\right)}={u}\cdot{u}+{u}\cdot{1}+{1}\cdot{u}+{1}\cdot{1}$

$\displaystyle={{u}}^{{2}}+{u}+{u}+{1}$

$\displaystyle={{u}}^{{2}}+{2}{u}+{1}$

Then let $\displaystyle{\left({u}+{1}\right)}{\left({u}+{1}\right)}={{u}}^{{2}}+{2}{u}+{1}\in{\left({u}+{1}\right)}{\left({u}+{1}\right)}{\left({u}+{1}\right)}$ :

$\displaystyle{\left({u}+{1}\right)}{\left({u}+{1}\right)}{\left({u}+{1}\right)}={\left({u}+{1}\right)}\cdot{\left({{u}}^{{2}}+{2}{u}+{1}\right)}$

Applying distributive property:

$\displaystyle{\left({u}+{1}\right)}{\left({{u}}^{{2}}+{2}{u}+{1}\right)}={u}\cdot{\left({{u}}^{{2}}+{2}{u}+{1}\right)}+{1}\cdot{\left({{u}}^{{2}}+{2}{u}+{1}\right)}$

$\displaystyle={{u}}^{{3}}+{2}{{u}}^{{2}}+{u}+{{u}}^{{2}}+{2}{u}+{1}$

$\displaystyle={{u}}^{{3}}+{3}{{u}}^{{2}}+{3}{u}+{1}$

then $\displaystyle{x}={\left({u}+{1}\right)}\cdot{\left({u}+{1}\right)}\cdot{\left({u}+{1}\right)}$ is the same as

$\displaystyle{x}={{u}}^{{3}}+{3}{{u}}^{{2}}+{3}{u}+{1}$

Substitute $\displaystyle{x}={{u}}^{{3}}+{3}{{u}}^{{2}}+{3}{u}+{1}\in{3}\int\frac{{x}}{{u}}{d}{u}$ :

$\displaystyle{3}\int\frac{{x}}{{u}}{d}{u}={3}\int\frac{{{{u}}^{{3}}+{3}{{u}}^{{2}}+{3}{u}+{1}}}{{u}}{d}{u}$

$\displaystyle={3}\int{\left(\frac{{{u}}^{{3}}}{{u}}+\frac{{{3}{{u}}^{{2}}}}{{u}}+\frac{{{3}{u}}}{{u}}+\frac{{1}}{{u}}\right)}{d}{u}$

$\displaystyle={3}\int{\left({{u}}^{{2}}+{3}{u}+{3}+\frac{{1}}{{u}}\right)}{d}{u}$

Evaluating each term in separate integral:

$\displaystyle{3}\cdot{\left[\int{{u}}^{{2}}\cdot{d}{u}+\int{3}{u}\cdot{d}{u}+\int{3}\cdot{d}{u}+\int\frac{{1}}{{u}}{d}{u}\right]}$

where:

$\displaystyle\int{{u}}^{{2}}\cdot{d}{u}=\frac{{{u}}^{{3}}}{{3}}$

$\displaystyle\int{3}{u}\cdot{d}{u}=\frac{{{3}{{u}}^{{2}}}}{{2}}$

$\displaystyle\int{3}\cdot{d}{u}={3}{u}$

$\displaystyle\int\frac{{1}}{{u}}{d}{u}={\ln}{\left|{u}\right|}$

$\displaystyle{3}\cdot{\left[\int{{u}}^{{2}}\cdot{d}{u}+\int{3}{u}\cdot{d}{u}+\int{3}\cdot{d}{u}+\int\frac{{1}}{{u}}{d}{u}\right]}$ becomes:

$\displaystyle{3}\cdot{\left[\frac{{{u}}^{{3}}}{{3}}+\frac{{{3}{{u}}^{{2}}}}{{2}}+{3}{u}+{\ln}{\left|{u}\right|}\right]}+{C}={3}\frac{{{u}}^{{3}}}{{3}}+\frac{{{9}{{u}}^{{2}}}}{{2}}+{9}{u}+{3}{\ln}{\left|{u}\right|}+{C}$

Substitute u = root(3)(x)-1:

$\displaystyle{3}\frac{{{u}}^{{3}}}{{3}}+\frac{{{9}{{u}}^{{2}}}}{{2}}+{9}{u}+{3}{\ln}{\left|{u}\right|}+{C}={{\left({\sqrt[{{3}}]{{{x}}}}-{1}\right)}}^{{3}}+\frac{{{9}{{\left({\sqrt[{{3}}]{{{x}}}}-{1}\right)}}^{{2}}}}{{2}}+{9}{\left({\sqrt[{{3}}]{{{x}}}}-{1}\right)}+{3}{\ln}{\left|{\left({\sqrt[{{3}}]{{{x}}}}-{1}\right)}\right|}+{C}$

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$\displaystyle\int\frac{{\sqrt[{{3}}]{{{x}}}}}{{{\sqrt[{{3}}]{{{x}}}}-{1}}}{\left.{d}{x}\right.}$ Find the indefinite integral by u substitution. (let u be the denominator of the integral)

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Find the indefinite integral by u substitution. (let u be the denominator of the integral)

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$\displaystyle\int\frac{{\sqrt[{{3}}]{{{x}}}}}{{{\sqrt[{{3}}]{{{x}}}}-{1}}}{\left.{d}{x}\right.}$ Find the indefinite integral by u substitution. (let u be the denominator of the integral)

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